If `a=(sqrt(5)+1)/(sqrt(5)-1)` and `b=(sqrt(5)-1)/(sqrt(5)+1)` , the value of `((a^(2)+ab+b^(2))/(a^(2)-ab+b^(2)))` is

Options

A. `(3)/(4)`
B. `(4)/(3)`
C. `(3)/(5)`
D. `(5)/(3)`

Answer

To solve the expression \(\frac{a^2 + ab + b^2}{a^2 - ab + b^2}\) given \(a = \frac{\sqrt{5} + 1}{\sqrt{5} - 1}\) and \(b = \frac{\sqrt{5} - 1}{\sqrt{5} + 1}\), we will follow these steps:

### Step 1: Calculate \(a^2\) and \(b^2\)

First, we need to find \(a^2\) and \(b^2\):

\[ a^2 = \left(\frac{\sqrt{5} + 1}{\sqrt{5} - 1}\right)^2 = \frac{(\sqrt{5} + 1)^2}{(\sqrt{5} - 1)^2} = \frac{5 + 2\sqrt{5} + 1}{5 - 2\sqrt{5} + 1} = \frac{6 + 2\sqrt{5}}{6 - 2\sqrt{5}} \]

\[ b^2 = \left(\frac{\sqrt{5} - 1}{\sqrt{5} + 1}\right)^2 = \frac{(\sqrt{5} - 1)^2}{(\sqrt{5} + 1)^2} = \frac{5 - 2\sqrt{5} + 1}{5 + 2\sqrt{5} + 1} = \frac{6 - 2\sqrt{5}}{6 + 2\sqrt{5}} \]

### Step 2: Calculate \(ab\)

Next, we calculate \(ab\):

\[ ab = \left(\frac{\sqrt{5} + 1}{\sqrt{5} - 1}\right) \left(\frac{\sqrt{5} - 1}{\sqrt{5} + 1}\right) = \frac{(\sqrt{5} + 1)(\sqrt{5} - 1)}{(\sqrt{5} - 1)(\sqrt{5} + 1)} = \frac{5 - 1}{5 - 1} = 1 \]

### Step 3: Substitute into the expression

Now we substitute \(a^2\), \(b^2\), and \(ab\) into the expression \(\frac{a^2 + ab + b^2}{a^2 - ab + b^2}\):

\[ a^2 + b^2 = \frac{6 + 2\sqrt{5}}{6 - 2\sqrt{5}} + \frac{6 - 2\sqrt{5}}{6 + 2\sqrt{5}} \]

To combine these fractions, we need a common denominator:

\[ = \frac{(6 + 2\sqrt{5})^2 + (6 - 2\sqrt{5})^2}{(6 - 2\sqrt{5})(6 + 2\sqrt{5})} \]

Calculating the numerator:

\[ (6 + 2\sqrt{5})^2 = 36 + 24\sqrt{5} + 20 = 56 + 24\sqrt{5} \]
\[ (6 - 2\sqrt{5})^2 = 36 - 24\sqrt{5} + 20 = 56 - 24\sqrt{5} \]

Adding these gives:

\[ (56 + 24\sqrt{5}) + (56 - 24\sqrt{5}) = 112 \]

The denominator simplifies to:

\[ (6 - 2\sqrt{5})(6 + 2\sqrt{5}) = 36 - 20 = 16 \]

Thus:

\[ a^2 + b^2 = \frac{112}{16} = 7 \]

Now substituting back into the expression:

\[ \frac{a^2 + ab + b^2}{a^2 - ab + b^2} = \frac{7 + 1}{7 - 1} = \frac{8}{6} = \frac{4}{3} \]

### Final Answer

The value of \(\frac{a^2 + ab + b^2}{a^2 - ab + b^2}\) is \(\frac{4}{3}\).